• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

The relationships between the number of different spacers in an arrangement of square tiles and the dimensions of the tiles in the same arrangement.

Extracts from this document...

Introduction

image01.png

Introduction

We have been given the task of investigating the relationships between the number of different spacers in an arrangement of square tiles and the dimensions of the tiles in the same arrangement. I will begin my investigation by researching square arrangements of tiles, and then move onto rectangular arrangements. I will then investigate triangle arrangements.

Stage 1 – Square arrangements

The spacers that will be used in this investigation are –

+ Spacer        T Spacer        L Spacer

  I began by drawing 5 different arrangements of tiles, beginning with a 1x1 arrangement, and finishing with a 5x5 arrangement. I drew these on a separate piece of graph paper. (See sheet S1).

  The results gathered from these sketches are shown here:

Pattern number

Number of Squares

+ Spacers

T Spacers

L Spacers

1

1

0

0

4

2

4

1

4

4

3

9

4

8

4

4

16

9

12

4

5

25

16

16

4

...read more.

Middle

  I have now drawn the 10x10 arrangement and have written the actual properties as read from my sketch in this table:

Pattern number

Number of Squares

+ Spacers

T Spacers

L Spacers

10

100

81

36

4

  From the results gathered by actually drawing the arrangement I have deduced that the rules that I obtained from my original set of results are in fact correct for the square arrangements that I have drawn so far, and they appear to be able to hold true when tested with larger arrangements.  I have therefore decided to move onto the next stage of my investigation.

Stage 2 – Rectangular arrangements

For the next stage of my investigation, I will investigate the rules for spacers when used with rectangular arrangements. These rules will apply to any arrangement of squares, as long as they are in a rectangle. Therefore I do not need to draw out a table like the squares section.

  The rule for the number of squares in the squares section, xy, will apply to rectangles as well.

...read more.

Conclusion

  When investigating the rule for the * shaped spacers, I came across a problem. I tried to use quadrilateral equations to find a rule, but this didn’t work. So I decided to approach the problem from a different perspective. Firstly, I doubled the amount of * spacers in a triangle, thus forming a rectangle. (See sheet T2). Now I could apply the rule I used to find the amount of squares in a rectangle, which is xy. I saw that x is always two less than the pattern number (n), and y is always one less. So I made the rule (n-2) x (n-1). But this was always double the amount of the * shaped spacers. Dividing the rule by two easily solved this. So I came out with the rule

(n-2) x (n-1)image00.png

         2

I simplified this to n  - 3n + 1

                         2

I now feel that I have fulfilled my original coursework assignment and do not have to investigate further. I have encountered some problems, which I overcame.

Name: Sam Koprowski

Candidate Number: 7393

School: George Ward

Centre Number: 66633

...read more.

This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Emma's Dilemma essays

  1. Emma's Dilemma

    of = ( Number of Y's + 1 ) combinations For example: To find the number of combinations which can be made from using 10 letter Y's and One X, can be found by using the following formula: No. of = 10 + 1 Combinations = 11 Two X's and an increasing number of Y's Two X's and No

  2. I am doing an investigation into words and their number of combinations. I will ...

    After a while I found that if I did the basic formula-n!/a! (Presuming a is the number of letters the same) I got twice the amount I had for the combination. From there it was quite simple-I tried doing (n!/a!)/2 and got the right amount for the combinations.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work